hdu 4602 Partition（矩阵快速幂乘法）

Problem Description

Define f(n) as the number of ways to perform n in format of the sum of some positive integers. For instance, when n=4, we have
  4=1+1+1+1
  4=1+1+2
  4=1+2+1
  4=2+1+1
  4=1+3
  4=2+2
  4=3+1
  4=4
totally 8 ways. Actually, we will have f(n)=2(n-1) after observations.
Given a pair of integers n and k, your task is to figure out how many times that the integer k occurs in such 2(n-1) ways. In the example above, number 1 occurs for 12 times, while number 4 only occurs once.

 

Input

The first line contains a single integer T(1≤T≤10000), indicating the number of test cases.
Each test case contains two integers n and k(1≤n,k≤109).

 

 

Output

Output the required answer modulo 109+7 for each test case, one per line.

 

Sample Input

2
4 2
5 5

 

Sample Output

5 
1

 

 

 

Source

2013 Multi-University Training Contest 1

 

题目大意：将一个数 n 拆分，问所有的拆分组合中 K 出现了几次。

思路：

列出了 n=5 时 5,4,3,2,1 出现的次数为 1 2 5 12 28
f[n+1]=3*f[n]-f[n-1]-f[n-2]-..f[1]
f[n]=3*f[n-1]-f[n-2]-..f[1]
==> f[n+1]=4*f[n]-4*f[n-1]

 

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<math.h>
#include<algorithm>
#include<queue>
#include<set>
#include<bitset>
#include<map>
#include<vector>
#include<stdlib.h>
#include <stack>
using namespace std;
#define PI acos(-1.0)
#define max(a,b) (a) > (b) ? (a) : (b)
#define min(a,b) (a) < (b) ? (a) : (b)
#define ll long long
#define eps 1e-10
#define MOD 1000000007
#define N 1000000
#define inf 1e12
ll n,k;
struct Matrix{
   ll mp[3][3];
};
Matrix Mul(Matrix a,Matrix b){
   Matrix res;
   for(ll i=0;i<2;i++){
      for(ll j=0;j<2;j++){
         res.mp[i][j]=0;
         for(ll k=0;k<2;k++){
            res.mp[i][j]=(res.mp[i][j]+(a.mp[i][k]*b.mp[k][j])%MOD+MOD)%MOD;
         }
      }
   }
   return res;
}
Matrix fastm(Matrix a,ll b){
   Matrix res;
   memset(res.mp,0,sizeof(res.mp));
   for(ll i=0;i<2;i++){
      res.mp[i][i]=1;
   }
   while(b){
      if(b&1){
         res=Mul(res,a);
      }
      a=Mul(a,a);
      b>>=1;
   }
   return res;
}
int main()
{
   ll t;
   scanf("%I64d",&t);
   while(t--){
      scanf("%I64d%I64d",&n,&k);
      if(k>n){
         printf("0\n");
         continue;
      }
      ll tmp=n-k+1;
      if(tmp==1){
         printf("1\n");
         continue;
      }
      if(tmp==2){
         printf("2\n");
         continue;
      }
      if(tmp==3){
         printf("5\n");
         continue;
      }

      Matrix ttt;
      ttt.mp[0][0]=4;
      ttt.mp[0][1]=-4;
      ttt.mp[1][0]=1;
      ttt.mp[1][1]=0;

      ttt=fastm(ttt,tmp-3);

      Matrix cnt;
      cnt.mp[0][0]=5;
      cnt.mp[1][0]=2;
      ttt=Mul(ttt,cnt);
      printf("%I64d\n",ttt.mp[0][0]);

   }
    return 0;
}